【工具】Latex 公式用法积累

野性酷女 2023-08-17 17:13 116阅读 0赞

公式更建议用 **Mathpix snipping tool** 这个软件识别哈，没办法才看我这个哈

∣ λ − a 11 − a 12 … − a 1 n − a 21 λ − a 22 … − a 2 n … … … … − a n 1 − a n 2 … λ − a n n ∣ \\left| \\begin\{array\}\{cccc\} \{\\lambda-a\_\{11\}\} & \{-a\_\{12\}\} & \{\\dots\} & \{-a\_\{1 n\}\} \\\\ \{-a\_\{21\}\} & \{\\lambda-a\_\{22\}\} & \{\\dots\} & \{-a\_\{2 n\}\} \\\\ \{\\dots\} & \{\\dots\} & \{\\dots\} & \{\\dots\} \\\\ \{-a\_\{n 1\}\} & \{-a\_\{n 2\}\} & \{\\dots\} & \{\\lambda-a\_\{n n\}\} \\end\{array\} \\right| ∣∣∣∣∣∣∣∣λ−a11−a21…−an1−a12λ−a22…−an2…………−a1n−a2n…λ−ann∣∣∣∣∣∣∣∣

\left|
    \begin{ array}{ cccc}
    { \lambda-a_{ 11}} & { -a_{ 12}} & { \dots} & { -a_{ 1 n}} \\ 
    { -a_{ 21}} & { \lambda-a_{ 22}} & { \dots} & { -a_{ 2 n}} \\ 
    { \dots} & { \dots} & { \dots} & { \dots} \\ 
    { -a_{ n 1}} & { -a_{ n 2}} & { \dots} & { \lambda-a_{ n n}}
    \end{ array}
    \right|

y = \[ y 1 ⋮ y n \] y = \\left\[ \\begin\{array\}\{c\} y\_1 \\\\ \\vdots \\\\ y\_n \\end\{array\} \\right\] y=⎣⎢⎡y1⋮yn⎦⎥⎤

y = 
    \left[
    \begin{ array}{ c} 
    y_1 \\ \vdots \\ y_n
    \end{ array}
    \right]

y = 1 = 2 \\begin\{aligned\} y & = 1 \\\\ & = 2 \\end\{aligned\} y=1=2

\begin{ aligned}
    y
    & = 1 \\
    & = 2
    \end{ aligned}

\{ v 1 = 1 v 2 = − 1 v 3 = − 1 \\left\\\{ \\begin\{array\}\{l\} \{v\_1 = 1\} \\\\ \{v\_2 = -1\} \\\\ \{v\_3 = -1\} \\end\{array\} \\right. ⎩⎨⎧v1=1v2=−1v3=−1

\left\{ 
    \begin{ array}{ l}
    v_1 = 1 \\ 
    v_2 = -1 \\
    v_3 = -1
    \end{ array}
    \right.

\{ p , x = 1 1 − p , x = 0 \\begin\{cases\} p, & x = 1 \\\\ 1 - p, & x = 0 \\end\{cases\} \{ p,1−p,x=1x=0

$$
    \begin{ cases}
      p, & x = 1 \\
      1 - p, & x = 0
    \end{ cases}
    $$

## 箭头 ##

<table> 
 <thead> 
 <tr> 
 <th>效果</th> 
 <th>写法</th> 
 </tr> 
 </thead> 
 <tbody> 
 <tr> 
 <td> → \rightarrow →</td> 
 <td><code>\rightarrow</code></td> 
 </tr> 
 <tr> 
 <td> ⇒ \Rightarrow ⇒</td> 
 <td><code>\Rightarrow</code></td> 
 </tr> 
 <tr> 
 <td> ↔ \leftrightarrow ↔</td> 
 <td><code>\leftrightarrow</code></td> 
 </tr> 
 <tr> 
 <td> ⟶ \longrightarrow ⟶</td> 
 <td><code>\longrightarrow</code></td> 
 </tr> 
 </tbody> 
</table>

## 空格 ##

<table> 
 <thead> 
 <tr> 
 <th>效果</th> 
 <th>写法</th> 
 </tr> 
 </thead> 
 <tbody> 
 <tr> 
 <td> a b a \qquad b ab</td> 
 <td><code>a \qquad b</code></td> 
 </tr> 
 <tr> 
 <td> a b a \quad b ab</td> 
 <td><code>a \quad b</code></td> 
 </tr> 
 <tr> 
 <td> a &nbsp; b a \ b a&nbsp;b</td> 
 <td><code>a \ b</code></td> 
 </tr> 
 </tbody> 
</table>

## 符号 ##

<table> 
 <thead> 
 <tr> 
 <th>效果</th> 
 <th>写法</th> 
 </tr> 
 </thead> 
 <tbody> 
 <tr> 
 <td> ≥ \geq ≥</td> 
 <td><code>$\geq$</code></td> 
 </tr> 
 <tr> 
 <td> ≤ \leq ≤</td> 
 <td><code>$\leq$</code></td> 
 </tr> 
 <tr> 
 <td> ≠ \neq =</td> 
 <td><code>$\neq$</code></td> 
 </tr> 
 <tr> 
 <td> ∝ \propto ∝</td> 
 <td><code>\propto</code></td> 
 </tr> 
 <tr> 
 <td> ∀ \forall ∀</td> 
 <td><code>\forall</code></td> 
 </tr> 
 <tr> 
 <td> ∃ \exists ∃</td> 
 <td><code>\exists</code></td> 
 </tr> 
 <tr> 
 <td> a ^ b c \hat abc a^bc</td> 
 <td><code>$\hat a$</code></td> 
 </tr> 
 <tr> 
 <td> a b c ^ \widehat{abc} abc </td> 
 <td><code>$\widehat{a}$</code></td> 
 </tr> 
 <tr> 
 <td> ≜ \triangleq ≜</td> 
 <td><code>$\triangleq$</code></td> 
 </tr> 
 <tr> 
 <td> ∑ i = 1 \sum_{i=1} ∑i=1</td> 
 <td><code>$\sum_{i=1}$</code></td> 
 </tr> 
 <tr> 
 <td> ∑ i = 1 \sum\limits_{i=1} i=1∑</td> 
 <td><code>$\sum\limits_{i=1}$</code></td> 
 </tr> 
 <tr> 
 <td> ∏ i = 1 \prod_{i=1} ∏i=1</td> 
 <td><code>$\prod_{i=1}$</code></td> 
 </tr> 
 <tr> 
 <td> ∏ i = 1 \prod\limits_{i=1} i=1∏</td> 
 <td><code>$\prod\limits_{i=1}$</code></td> 
 </tr> 
 <tr> 
 <td> arg ⁡ min ⁡ 0 ≤ j ≤ k − 1 \underset{0\leq j \leq k-1}{\arg\min} 0≤j≤k−1argmin</td> 
 <td><code>$\underset{0\leq j \leq k-1}{\arg\min}$</code></td> 
 </tr> 
 <tr> 
 <td> B A \underset{A}{B} AB</td> 
 <td><code>$\underset{A}{B}$</code></td> 
 </tr> 
 <tr> 
 <td> a = ? b a\overset{?}=b a=?b</td> 
 <td><code>$a \overset{?}= b$</code></td> 
 </tr> 
 <tr> 
 <td> ∑ 0 &lt; i ≤ n 0 &lt; j ≤ n A i j \sum_{0&lt;i \leq n \atop 0&lt;j \leq n} A_{i j} ∑0&lt;j≤n0&lt;i≤nAij</td> 
 <td><code>$\sum_{0&lt;i \leq n \atop 0&lt;j \leq n} A_{i j}$</code></td> 
 </tr> 
 <tr> 
 <td> ∈ \in ∈</td> 
 <td><code>$\in$</code></td> 
 </tr> 
 <tr> 
 <td> ∉ \notin ∈/</td> 
 <td><code>$\notin$</code></td> 
 </tr> 
 <tr> 
 <td> A B C D E F G H I J K L M N O P Q R S T U V W X Y Z \mathcal{ABCDEFGHIJKLMNOPQRSTUVWXYZ} ABCDEFGHIJKLMNOPQRSTUVWXYZ</td> 
 <td><code>$\mathcal{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$</code></td> 
 </tr> 
 </tbody> 
</table>

多行公式对齐  
可以控制左右的对齐方式  
 max ⁡ g ( λ , ν ) subject to λ ≥ 0 \\begin\{array\}\{cl\} \\max & g(\\lambda, \\nu) \\\\ \\text \{subject to\} & \\lambda \\ge 0 \\end\{array\} maxsubject tog(λ,ν)λ≥0

$$
    \begin{array}{cl}
    \max & g(\lambda, \nu) \\
    \text { subject to } & \lambda \ge 0
    \end{array}
    $$

max ⁡ g ( λ , ν ) subject to λ ≥ 0 \\begin\{array\}\{ll\} \\max & g(\\lambda, \\nu) \\\\ \\text \{subject to\} & \\lambda \\ge 0 \\end\{array\} maxsubject tog(λ,ν)λ≥0

$$
    \begin{array}{ll}
    \max & g(\lambda, \nu) \\
    \text { subject to } & \lambda \ge 0
    \end{array}
    $$

## 参考 ##

http://blog.lisp4fun.com/2017/11/01/formula